Reused writeup on Math-6117 (gpt pro 5.2)

Erdős Problem 1063 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1063

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1063.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1063.tex.

FORMAL RESTATEMENT

Fix an integer $k\ge 2$. For each integer $n\ge 2k$ consider the $k$ integers \[ n,\ n-1,\ \dots,\ n-(k-1). \] Let $D(n,k)$ be the number of indices $i\in\{0,1,\dots,k-1\}$ such that $(n-i)$ divides $\binom{n}{k}$. Define $n_k$ to be the least $n\ge 2k$ for which $D(n,k)=k-1$ (i.e. all but one of these $k$ numbers divide $\binom{n}{k}$). The task is to estimate $n_k$.

QUICK LITERATURE/CONTEXT CHECK

The problem statement records several facts: $1$ for $n\ge 2k$ at least one $n-i$ fails to divide $\binom{n}{k}$, $2$ $n_2=4,n_3=6,n_4=9,n_5=12$, and $3$ an upper bound $n_k\le k!$ for $k\ge 3$ (and a refinement using an lcm). I prove the $k!$ upper bound from scratch below.

ATTACK PLAN

Proof-track: prove explicit general upper bounds by exhibiting a concrete $n$ (in terms of $k$) with $D(n,k)=k-1$. Computation-track: directly search for $n_k$ for small $k$ to understand size/growth.

WORK

FAST REALITY CHECK. A direct search over $n\ge 2k$ (using exact integer binomials) gives:

k : n_k (and the unique failing i)
 2 : 4    (fails i=0)
 3 : 6    (fails i=0)
 4 : 9    (fails i=1)
 5 : 12   (fails i=2)
 6 : 75   (fails i=3)
 7 : 30   (fails i=2)
 8 : 70   (fails i=6)
 9 : 56   (fails i=2)
10 : 2403 (fails i=3)
11 : 280  (fails i=5)
12 : 3465 (fails i=9)
13 : 210  (fails i=2)
14 : 793  (fails i=9)
15 : 4732 (fails i=7)

The first four values match the ones stated in the problem text.

Lemma 1063.1 (a sufficient condition for $n-i\mid \binom{n}{k}$ when $i\ge 1$). Let $k\ge 2$ and $n\ge k$. If $k!\mid n$, then for every $i\in\{1,2,\dots,k-1\}$ we have \[ (n-i)\ \Big|\ \binom{n}{k}. \]

Proof. For $1\le i\le k-1$, \[ \frac{\binom{n}{k}}{n-i} = \frac{n(n-1)\cdots(n-k+1)}{k!\,(n-i)} = \frac{n\cdot\prod_{\substack{1\le j\le k-1\\ j\ne i}} (n-j)}{k!}. \] If $k!\mid n$, write $n=k!\,t$ with $t\in\mathbb Z_{\ge 1}$. Then the right-hand side equals \[ t\cdot\prod_{\substack{1\le j\le k-1\\ j\ne i}} (n-j), \] which is an integer. Hence $\binom{n}{k}/(n-i)\in\mathbb Z$, i.e. $(n-i)\mid \binom{n}{k}$. $\square$

Lemma 1063.2 ($n=k!$ makes $i=0$ fail). For every $k\ge 2$, with $n:=k!$ we have \[ n\nmid \binom{n}{k}. \]

Proof. We have \[ \frac{\binom{n}{k}}{n} = \frac{(n-1)(n-2)\cdots (n-k+1)}{k!}. \] Let $P:=\prod_{j=1}^{k-1} (n-j)$. Reducing modulo $k!=n$ gives \[ P \equiv \prod_{j=1}^{k-1} (-j) = (-1)^{k-1}(k-1)! \pmod{n}. \] Since $0<(k-1)!<k!=n$, the residue $(-1)^{k-1}(k-1)!$ is nonzero modulo $n$, so $n\nmid P$. Therefore $k!=n$ does not divide $P$, and $\binom{n}{k}/n$ is not an integer. Equivalently $n\nmid \binom{n}{k}$. $\square$

Corollary 1063.3 (general upper bound $n_k\le k!$ for $k\ge 3$). For every $k\ge 3$, \[ n_k \le k!. \]

Proof. Let $n=k!$. Then $n\ge 2k$ for $k\ge 3$. By Lemma 1063.1 (since $k!\mid n$), we have $(n-i)\mid \binom{n}{k}$ for every $i=1,2,\dots,k-1$. By Lemma 1063.2, $n$ itself (i.e. $i=0$) does not divide $\binom{n}{k}$. Thus $D(n,k)=k-1$, so by definition $n_k\le n=k!$. $\square$

VERIFICATION

Lemma 1063.1 explicitly rewrites $\binom{n}{k}/(n-i)$ and uses only the divisibility $k!\mid n$. Lemma 1063.2 checks the divisibility failure by a clean congruence modulo $n=k!$; the key point is that $(-1)^{k-1}(k-1)!$ is nonzero modulo $k!$. The computed values of $n_k$ for $k\le 15$ were obtained by scanning $n$ upward from $2k$ and checking all $k$ divisibilities.

FINAL

UNRESOLVED

1. (i) Strongest proved partial result. A fully explicit construction gives $n_k\le k!$ for all $k\ge 3$ (Corollary 1063.3). Exact computation yields $n_2=4,n_3=6,n_4=9,n_5=12$ and additional values listed in the FAST REALITY CHECK up to $k=15$.
2. (ii) First gap (crisp). Determine the correct order of growth of $n_k$ as $k\to\infty$ (e.g. polynomial, exponential, factorial) and obtain matching lower bounds.
3. (iii) Top 3 next moves. $1$ Try to prove the stronger explicit upper bound mentioned in the problem statement, $n_k\le k\cdot \mathrm{lcm}(2,3,\dots,k-1)$, by adapting Lemma 1063.1 to weaker divisibility hypotheses than $k!\mid n$. $2$ Develop lower bounds on $n_k$ by showing that for $n$ below a certain threshold, at least two of the $k$ numbers $n-i$ must fail to divide $\binom{n}{k}$. $3$ Extend computations of $n_k$ to larger $k$ to guess the typical size and whether $n_k$ is often much smaller than $k!$.
4. (iv) Minimal counterexample structure. A counterexample to an anticipated growth rate (say $n_k\le \exp(Ck)$) would require $k$ for which every $n$ up to that scale has at least two indices $i$ with $(n-i)\nmid \binom{n}{k}$. Such a $k$ might force many small prime-power obstructions simultaneously.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.